Six Degrees of Cowvin Bacon POJ - 2139

The cows have been making movies lately, so they are ready to play a variant of the famous game "Six Degrees of Kevin Bacon". 

The game works like this: each cow is considered to be zero degrees of separation (degrees) away from herself. If two distinct cows have been in a movie together, each is considered to be one 'degree' away from the other. If a two cows have never worked together but have both worked with a third cow, they are considered to be two 'degrees' away from each other (counted as: one degree to the cow they've worked with and one more to the other cow). This scales to the general case. 

The N (2 <= N <= 300) cows are interested in figuring out which cow has the smallest average degree of separation from all the other cows. excluding herself of course. The cows have made M (1 <= M <= 10000) movies and it is guaranteed that some relationship path exists between every pair of cows. 

Input

* Line 1: Two space-separated integers: N and M 

* Lines 2..M+1: Each input line contains a set of two or more space-separated integers that describes the cows appearing in a single movie. The first integer is the number of cows participating in the described movie, (e.g., Mi); the subsequent Mi integers tell which cows were. 

Output

* Line 1: A single integer that is 100 times the shortest mean degree of separation of any of the cows. 

Sample Input

4 2
3 1 2 3
2 3 4

Sample Output

100

Hint

[Cow 3 has worked with all the other cows and thus has degrees of separation: 1, 1, and 1 -- a mean of 1.00 .] 

#include <iostream>
#include <cstdio>

using namespace std;

typedef long long ll;

const int M=300+5;
const int INF=1000;

int a[M][M];
int b[M];

void draw(int x){
    for(int i=1;i<=x;i++){
        for(int j=1;j<=x;j++){
            printf("%5d",a[i][j]);
        }
        cout<<endl;
    }
}

void solve(){
    int n,m;
    cin>>n>>m;

    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(i==j){
                a[i][j]=0;
            }
            else{
                a[i][j]=INF;
            }
        }
    }

    //draw(n);

    while(m--){
        int x;
        cin>>x;
        for(int i=1;i<=x;i++){
            cin>>b[i];
        }

        for(int i=1;i<x;i++){
            for(int j=i+1;j<=x;j++){
                a[b[i]][b[j]]=1;
                a[b[j]][b[i]]=1;
            }
        }

    }

     for(int k=1;k<=n;k++)
            for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++)a[i][j]=min(a[i][j],a[i][k]+a[k][j]);

   // draw(n);
    float minn=INF*INF;
    for(int i=1;i<=n;i++){

        float now=0;
        for(int j=1;j<=n;j++){
            now+=a[i][j];
        }
        minn=min(minn,now);
    }
    cout<<(int)(minn*100/(n-1))<<endl;

}

int main(){
    solve();
    return 0;
}